[bergkamp10]

Hi all, I'm obviously new here so I hope I'm not enacting any forum social faux pas. I just have a few questions about calcium fluoride or calcium difluoride (I'm not sure what is the convention of naming here )

I was looking at my lecture notes from last year and it gives this chemical equation in the context of saliva:

At [F⁺] > 100ppm and pH < 5; Ca₁₀(PO₄)₆(OH)₂ + 20F⁺ 10CaF₂ + 6HPO₄^2- + 2OH₂

i.e. Hydroxyapatite + Fluoride   Calcium Fluoride + Phosphoric Acid + Hydroxide Ion

Firstly the charge of fluoride and the balance of the hydroxide are both incorrect so I am a bit skeptical of the correctness of the rest of it. I know I don't have to know to even this depth of chemistry to pass my course but I find it a bit irritating when I don't fully understand something. I also find things harder to learn if I don't know th underlying science behind it e.g I often find that applying general chemistry, physics or calculus (I'm talking high school level here) really helps me understand things better.

Here are my queries:
1. I've written out this equation, Ca₁₀(PO₄)₆(OH)_2(s) + 20F^-_(aq) 10CaF_2(?) + 6HPO₄^2-_(aq) + 2OH^-_(aq) and I tried putting in the states but realised I was confused on the state of calcium fluoride and also where did the H come from for the HPO₄^-.

Now I'm pretty sure calcium fluoride is a solid but it is supposed to be an ion reservoir where it stays in saliva and dissolves at the optimal time (during acidic attack) but unfortunately my limited knowledge of chemistry confuses me here. If it is in saliva, does that mean it is aqueous or can a "solid" be solid chemically but not as most lay people think, and it is not necessary to be able to see them but molecularly, they are a solid structure? Perhaps it is soluble in saliva?

2. The notes and theory suggest that at a high concention of fluoride and and at lower pH which they give as pH < 5, the formation of calcium flouride is encouraged but then it also goes on to state that during acid attack on the oral environment, the calcium flouride dissolves and and provides the necessary ions for remineralisation of tooth tissue. I seem to find this notion hard to understand. Would acid attack not cause the pH to be low thus promoting calcium flouride formation and not it's dissolution into calcium and fluoride ions? or is this dissolution just a matter of need.

No matter how I look at it, adding more hydronium ions to the equation would cause the equalibirum to shift to formation of calcium fluoride if my understanding of Le Chatelier's Principle is correct thus their notion that low pH encourages the formation of calcium fluoride is correct but I can't fathom how it dissolutes during acidic attack?

Sorry for the long winded question, I look forward to being enlightened by the bright minds here. Thanks in advance!

[Abstractineum]

Hi bergkamp, I´m obviously new here too, but I´ll do my best answering your question:

First, I agree that the charge on the Flouride (or rather, the sign of the charge) is wrong. But regarding the balance of the hydroxide ions: On the left side of the equation you have got two (OH)^- ions, with the charge being excluded in the molecule formula. On the right side you have got two H₂O ions (you have written them as OH₂, which may have confused you). The difference is two H⁺ ions. Therefore, that which is unbalanced is the hydrogen ions. Same case with the HPO₄^2-.
  You could assume that they left the hydrogen ions out of the formula, as it is supposed to occur in an acid environment.

As with the state of Cacium Flouride and the remineralisation of tooth issue, the salt should be mostly solid in water. I agree with your conclusion, maybe someone else have an explanation?
  Your understanding of Le Chatelier´s principle is correct, but there may be another reaction taking place catalysed by the hydroxonium or by some enzyme.

Hope I managed to explain at least something 

[Borek]

Obviously you need water to balance the reaction:

Ca₁₀(PO₄)₆(OH)₂ + 20F^- + 6H₂O -> 10CaF₂ + 6HPO₄^2- + 8OH^-

However, this is overall reaction, and the real process goes through many stages and many equilibria are involved. See examples in the attached picture - and bear in mind these are only those I found in a limited database of one of the programs I have here.

Also remember that HF is a weak acid, so in low pH F^- will be converted to HF, shifting CaF₂ dissolution right. At pH 3.1 ratio [F^-]/[HF] equals 1, at pH 4.1 it equals 10 and so on.

[Abstractineum]

But as one of the conditions was pH<5, shouldn't it be H₃O⁺ on the left hand side and water on the right instead of water on the left and OH^- on the right? The results would be the same, just that it takes hydroxonium ions instead of produces hydroxide ones. Wouldn't that be preferable?

[bergkamp10]

Firstly, I just wanted to thank you both for helping me out. I've been studying for exams and this one slide has got me all bothered. I event went from my house to the library this morning and looked up the chapter and it has the same equation as in the lecture notes except without the mention of lower than pH 5 so perhaps that was a typo or mistake by the lecturer.

But as one of the conditions was pH<5, shouldn't it be H₃O⁺ on the left hand side and water on the right instead of water on the left and OH^- on the right? The results would be the same, just that it takes hydroxonium ions instead of produces hydroxide ones. Wouldn't that be preferable?

Hmm interesting...I've sought of given up as it was stalling my studying but I think I might get out some scrap paper and write some of them out. Nothing beats a piece of paper and pen sometimes.

I hope to talk to you guys more often, my exposure to chemistry has been unfortunately low since high school, I've forgotten so many things. Just today I was looking at the fundamental particles of physics and lamented how much I've fogotton.

Obviously you need water to balance the reaction:

Ca₁₀(PO₄)₆(OH)₂ + 20F^- + 6H₂O -> 10CaF₂ + 6HPO₄^2- + 8OH^-

However, this is overall reaction, and the real process goes through many stages and many equilibria are involved. See examples in the attached picture - and bear in mind these are only those I found in a limited database of one of the programs I have here.

Also remember that HF is a weak acid, so in low pH F^- will be converted to HF, shifting CaF₂ dissolution right. At pH 3.1 ratio [F^-]/[HF] equals 1, at pH 4.1 it equals 10 and so on.

Wow, adding that water seems to help make sense of it, I did think to add that but the more I thought about it, I got confused. I'm sure this is high school level and is quite embarassing to admit but I thought I was wrong to add water as I assumed it was thought the reaction was happening in water. Hmm...sometimes reasoning something out makes me more confused. Thank you so much though!

Hmmm well now I've tried both ways, adding either a water molecule or hydronium works too. A little knowledge begets more questions.

Side Note: Regarding the state of calcium fluoride, I believe the textbook I read at the library today mentions that it becomes bound to the tooth surface so I would guess that it is a solid that sort of just sticks to the tooth surface.

[Borek]

But as one of the conditions was pH<5, shouldn't it be H₃O⁺ on the left hand side and water on the right instead of water on the left and OH^- on the right? The results would be the same, just that it takes hydroxonium ions instead of produces hydroxide ones. Wouldn't that be preferable?

Just a matter of notation, as long as you are aware that the presence of water means also presence of OH^- and H⁺ you can write the equation any way you like. Using H⁺ in acidic solutions is a good idea.

Hmmm well now I've tried both ways, adding either a water molecule or hydronium works too. A little knowledge begets more questions.

As I wrote above - presence of water implies presence of both OH^- and H⁺:

H₂O <-> H⁺ + OH^-

and in all cases when you have to balance reaction equation you can to some extent use any combination of H₂O, H⁺ and OH^- that fits your needs. Of course using both H⁺ and OH^-  on one side doesn't make sense.

[bergkamp10]

As I wrote above - presence of water implies presence of both OH^- and H⁺:

H₂O <-> H⁺ + OH^-

and in all cases when you have to balance reaction equation you can to some extent use any combination of H₂O, H⁺ and OH^- that fits your needs. Of course using both H⁺ and OH^-  on one side doesn't make sense.

I see, wow that makes so much more sense! Thank you so much!